Calculus Examples

$\ln (e^{6}) - 2 \ln (e^{2}) = \log_{3} (x)$

Step 1

Rewrite the equation as $\log_{3} (x) = \ln (e^{6}) - 2 \ln (e^{2})$ .

$\log_{3} (x) = \ln (e^{6}) - 2 \ln (e^{2})$

Step 2

Simplify the right side.

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Step 2.1

Simplify $\ln (e^{6}) - 2 \ln (e^{2})$ .

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Step 2.1.1

Simplify each term.

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Step 2.1.1.1

Use logarithm rules to move $6$ out of the exponent.

$\log_{3} (x) = 6 \ln (e) - 2 \ln (e^{2})$

Step 2.1.1.2

The natural logarithm of $e$ is $1$ .

$\log_{3} (x) = 6 \cdot 1 - 2 \ln (e^{2})$

Step 2.1.1.3

Multiply $6$ by $1$ .

$\log_{3} (x) = 6 - 2 \ln (e^{2})$

Step 2.1.1.4

Use logarithm rules to move $2$ out of the exponent.

$\log_{3} (x) = 6 - 2 (2 \ln (e))$

Step 2.1.1.5

The natural logarithm of $e$ is $1$ .

$\log_{3} (x) = 6 - 2 (2 \cdot 1)$

Step 2.1.1.6

Multiply $2$ by $1$ .

$\log_{3} (x) = 6 - 2 \cdot 2$

Step 2.1.1.7

Multiply $- 2$ by $2$ .

$\log_{3} (x) = 6 - 4$

Step 2.1.2

Subtract $4$ from $6$ .

$\log_{3} (x) = 2$

Step 3

Rewrite $\log_{3} (x) = 2$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$3^{2} = x$

Step 4

Solve for $x$ .

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Step 4.1

Rewrite the equation as $x = 3^{2}$ .

$x = 3^{2}$

Step 4.2

Raise $3$ to the power of $2$ .

$x = 9$